Problem: Graph this system of equations and solve. $y = 2 x + 1$ $16x-4y = 12$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
Solution: The y-intercept for the first equation is $1$ , so the first line must pass through the point $(0, 1)$ The slope for the first equation is $2$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move up $1$ position to the right. $2$ positions up from $(0, 1)$ is $(1, 3)$ Graph the blue line so it passes through $(0, 1)$ and $(1, 3)$ Convert the second equation, $16x-4y = 12$ , to slope-intercept form. $y = 4 x - 3$ The y-intercept for the second equation is $-3$ , so the second line must pass through the point $(0, -3)$ The slope for the second equation is $4$ . Remember that the slope tells you rise over run. So in this case for every $4$ positions you move up $1$ position to the right. $4$ positions up from $(0, -3)$ is $(1, 1)$ Graph the green line so it passes through $(0, -3)$ and $(1, 1)$ The solution is the point where the two lines intersect. The lines intersect at $(2, 5)$.